Let $f: \mathbb{R} \to \mathbb{R}$ be a strictly monotone function satisfying $$ f(f(x)) - f(x) = e^x + x - 1 $$ for all $x \in \mathbb{R}$.
Can we conclude that $f$ must be decreasing?
(I don't know if there is such a function $f$, so the claim might be vacuously true.) We know that $f$ must be injective and $f(0) = 0$, so either $f(x) < 0$ for negative $x$ and $f(x) > 0$ for positive $x$ or the other way around. If we replaced $e^x + x - 1$ by $x$ then the answer would be no, as then $f(x) = \phi x$ would be an increasing solution, where $\phi$ is the golden ratio. Therefore a proof would have to use some property of $e^x + x - 1$ that $x$ alone doesn't have, if the claim is even true.
Thanks in advance for any help!